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Lec. 3Lec. 3Modeling of Dynamic System
2G 4G1G
4H
)(sY)(sR
3G
Quiz (10 mins)
Find the transfer function of the following block diagrams
2H
3H
1H
2
Solution:
2G 4G1G
4H)(sY
3G
)(sRA B
1
I1. Moving pickoff point A behind block
4G
H
1H
2H
3H4
1
G
4
1
G
4
3
G
H
4
2
G
H
3
2. Eliminate loop I and Simplify
II
443
432
1 HGG
GGG
+1G)(sY
B
H
)(sR
4
3
G
H
III
1H
4
2
G
H
II
332443
432
1 HGGHGG
GGG
++
I I I
4
142
G
HGH −
Not feedbackfeedback
4
)(sR )(sY
142
G
HGH −
332443
4321
1 HGGHGG
GGGG
++
3. Eliminate loop II & IIII
4G
143212321443332
4321
1 HGGGGHGGGHGGHGG
GGGG
sR
sY
−+++=)(
)(
5
• Static System: If a system does not change with time, it
is called a static system.
• The o/p depends on the i/p at the present time only
• Dynamic System: If a system changes with time, it is
called a dynamic system.
• The o/p depends on the i/p at the present time and the
past time. (integrator, differentiator).
6
• A system is said to be dynamic if its current outputmay depend on the past history as well as thepresent values of the input variables.
• Mathematically,
Time Input, ::
]),([)(
tu
tuty ≤≤= ττϕ 0
Time Input, :: tu
ExampleExampleExampleExample: A moving mass
MMMM
yyyyuuuu
ModelModelModelModel: Force=Mass XXXX Acceleration
uyM =ɺɺ
1. Mathematical.
2. State space.
9
A set of mathematical equations (e.g., differentialequs.) that describes the input-output behavior ofa system.
What is a model used for?What is a model used for?What is a model used for?What is a model used for?
• Simulation• Simulation
• Prediction/Forecasting
• Prognostics/Diagnostics
• Design/Performance Evaluation
• Control System Design
1. Electrical Systems1. Electrical Systems
Basic Elements of Electrical Systems
• The time domain expression relating voltage and current for the resistor is given by Ohm’s law i.e.
Rtitv RR )()( =
• The Laplace transform of the above equation is
RsIsV RR )()( =
Basic Elements of Electrical Systems
• The time domain expression relating voltage and current for the Capacitor is given as:
∫= 1dtti
Ctv cc ∫= )(
1)(
• The Laplace transform of the above equation (assuming there is no charge stored in the capacitor) is
)()( sICs
sV cc1=
Basic Elements of Electrical Systems
• The time domain expression relating voltage and current for the inductor is given as:
tdiLtv L )(
)( =dt
tdiLtv L
L)(
)( =
• The Laplace transform of the above equation (assuming there is no energy stored in inductor) is
)()( sLsIsV LL =
Component Symbol V-I Relation I-V Relation
Resistor Rtitv RR )()( =R
tvti R
R)(
)( =
15
Capacitor
Inductor
dt
tdiLtv L
L)(
)( =
dttiC
tv cc ∫= )()(1
dt
tdvCti c
c)(
)( =
dttvL
ti LL ∫= )()(1
The two-port network shown in the following figure has vi(t)as the input voltage and vo(t) as the output voltage. Find thetransfer function Vo(s)/Vi(s) of the network.
CCCCi(t)v ( t) v (t)
16
CCCCi(t)vi( t) vo(t)
∫+= dttiC
Rtitv i )()()(1
∫= dttiC
tv o )()(1
Taking Laplace transform of both equations, consideringinitial conditions to zero.
∫+= dttiC
Rtitv i )()()(1
∫= dttiC
tv o )()(1
)()()( sIRsIsV1+= )()( sIsV
1=
Re-arrange both equations as:
17
)()()( sICs
RsIsV i1+= )()( sI
CssV o
1=
)()( sIsCsV o =))(()(Cs
RsIsV i1+=
Substitute I(s) in equation on left
)()( sIsCsV o =))(()(Cs
RsIsV i1+=
))(()(Cs
RsCsVsV oi1+=
18
Cs
)()(
)(
CsRCssV
sV
i
o
11
+=
RCssV
sV
i
o
+=
1
1
)(
)(
The system has one pole at
RCssV
sV
i
o
+=
1
1
)(
)(
sRCs1
01 −=⇒=+
19
RCsRCs
101 −=⇒=+
Design an Electrical system that would place a pole at -4 ifadded to another system.
RCssV
sV
i
o
+=
1
1
)(
)(
CCCCi(t)vi( t) vo(t)
System has one pole at
Therefore,
20
RCs
1−=
14
RC− = − 250 1 mif R and C F= Ω =
iR(t)
+
IR(S)
+
TransformationTransformationTransformationTransformation
21
vR(t)
-
VR(S)
-
ZR = R
TransformationTransformationTransformationTransformation
iL(t)
vL(t)
+
IL(S)
VL(S)
+
ZL=LS
22
- -
ic(t)
vc(t)
+
Ic(S)
Vc(S)
+
ZC(S)=1/CS
23
- -
ZC(S)=1/CS
Consider following arrangement, find out equivalenttransform impedance.
L
CLRT ZZZZ ++=
24
C
R
CsLsRZT
1++=
CLRT ZZZZ
1111 ++=
CCCC
LLLL
CsLsRZ T 1
1111 ++=
25
CCCC
R
2. Mechanical Systems2. Mechanical Systems
• Part-I: Translational Mechanical System
• Part-II: Rotational Mechanical System
Mechanical Systems
• Part-III: Mechanical Linkages
27
Translational
Linear Motion
Rotational
Rotational Motion
28
Basic Elements of Translational Mechanical Systems
Translational SpringTranslational SpringTranslational SpringTranslational Spring
i)i)i)i)
Translational MassTranslational MassTranslational MassTranslational Mass
ii)ii)ii)ii)ii)ii)ii)ii)
Translational DamperTranslational DamperTranslational DamperTranslational Damper
iii)iii)iii)iii)
30
A translational spring is a mechanical element thatcan be deformed by an external force such that thedeformation is directly proportional to the forceapplied to it.
Translational SpringTranslational SpringTranslational SpringTranslational Spring
i)i)i)i)
Circuit SymbolsCircuit SymbolsCircuit SymbolsCircuit Symbols
31
Translational SpringTranslational SpringTranslational SpringTranslational Spring
i)i)i)i)
If F is the applied force
Then is the deformation if
Or is the deformation.
2x1x
02 =x1x
)( 21 xx −F
Or is the deformation.
The equation of motion is given as
Where is stiffness of spring expressed in N/m
32
)( 21 xx −
)( 21 xxkF −=
k
F
Translational Mass
Translational MassTranslational MassTranslational MassTranslational Mass
ii)ii)ii)ii)
• Translational Mass is aninertia element.
• A mechanical systemwithout mass does notexist.exist.
• If a force FFFF is applied to amass and it is displaced toxxxx meters then the relationb/w force anddisplacements is given byNewton’s law.
M)(tF
)(tx
xMF ɺɺ=33
Common Uses of DashpotsCommon Uses of DashpotsCommon Uses of DashpotsCommon Uses of DashpotsDoor Stoppers
Vehicle Suspension
Bridge SuspensionFlyover Suspension
34
Translational Damper
= −F = b x
• Where bbbb is damping coefficient (N/msN/msN/msN/ms----1111).
1 2( )F b x x= −
35
Example-1: Consider a simple horizontal spring-masssystem on a frictionless surface, as shown in figurebelow.
or
36
kxxm −=ɺɺ
0=+ kxxm ɺɺ
Consider the following system (friction is negligible)
k
F
xM
37
• Free Body Diagram
MF
kfMf
• Where and are force applied by the spring and inertial force respectively.
kf Mf
• Then the differential equation of the system is:
MF
kfMf
Mk ffF +=
38
• Then the differential equation of the system is:
kxxMF += ɺɺ
• Taking the Laplace Transform of both sides and ignoring initial conditions we get
)()()( skXsXMssF += 2
)()()( skXsXMssF += 2
• The transfer function of the system is
kMssF
sX
+=
2
1
)(
)(
Example-2
39
• if
12000
1000−=
=
Nmk
kgM
2
00102 +
=ssF
sX .
)(
)(
• The pole-zero map of the system is
2
00102 +
=ssF
sX .
)(
)(
Example-2
40Pole-Zero Map
40-1 -0.5 0 0.5 1-40
-30
-20
-10
0
10
20
30
Real Axis
Imag
inar
y A
xis
Consider the following system
k
F
2x
M1x B
41
• Mechanical Network
↑ M
k
BF
1x 2x
• Mechanical Network
↑ M
k
BF
1x 2x
42
)( 21 xxkF −=
At node 1x
At node 2x
22120 xBxMxxk ɺɺɺ ++−= )(
43
44
).()()( 10 eq =−+−+ ioioo xxkxxbxm ɺɺɺɺ
2 eq. iiooo kxxbkxxbxm +=++ ɺɺɺɺ
45
Taking Laplace Transform of the equation (2)
)()()()()( skXsbsXskXsbsXsXms iiooo +=++2
kbsms
kbs
sX
sX
i
o
+++=
2)(
)(
Car Body
46
Car Body
Bogie-2
Bogie
Frame
Bogie-1
WheelsetsPrimary
Suspension
Secondary
Suspension
47